A tuning fork of frequency 480 Hz produces 10 beats per second when sounded with a vibrating sonometer string. What must have been the frequency of the string if a slight increase in tension produces lesser beats per second than before
Text Solution
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If suppose n S = frequency of string 
n f = Frequency of tuning fork = 480 Hz
x = Beats heard per second = 10
as tension T increases, so n S increases ()
Also it is given that number of beats per sec decreases (i.e. x ↓ ↓ )
Hence n S – n f = x ↓ ↓ ... (i) Wrong
n f – n S = x ↓ ↓ ... (ii) Correct
⇒ ⇒ n S = n f – x = 480 – 10 = 470 Hz.
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